构造一个从x ^ 1，x ^ 2，……..，x ^ n获得的值的数字的频率数组

给定两个整数(x和n)。任务是找到一个数组，使其包含出现在(x ^ 1，x ^ 2，…。，x ^(n-1)，x ^(n))中的索引编号的频率。

例子：

Input: x = 15, n = 3
Output: 0 1 2 2 0 3 0 1 0 0
Numbers x^1 to x^n are 15, 225, 3375.
So frequency array is 0 1 2 2 0 3 0 1 0 0.

Input: x = 1, n = 5
Output: 0 5 0 0 0 0 0 0 0 0
Numbers x^1 to x^n are 1, 1, 1, 1, 1. 
So frequency of digits is 0 5 0 0 0 0 0 0 0 0.

方法：

1. 维护频率计数数组以存储数字0-9。
2. 遍历x ^ 1到x ^ n的每个数字，对于每个数字，在频率计数数组中的相应索引处加1。
3. 打印频率阵列

下面是上述方法的实现：

C++

// CPP implementation of above approach
#include
using namespace std;
  
// Function that traverses digits in a number and
// modifies frequency count array
void countDigits(double val, long arr[])
{
    while ((long)val > 0) {
        long digit = (long)val % 10;
        arr[(int)digit]++;
        val = (long)val / 10;
    }
    return;
}
  
void countFrequency(int x, int n)
{
  
    // Array to keep count of digits
    long freq_count[10]={0};
  
    // Traversing through x^1 to x^n
    for (int i = 1; i <= n; i++)
    {
        // For power function, both its parameters are
        // to be in double
        double val = pow((double)x, (double)i);
        // calling countDigits function on x^i
        countDigits(val, freq_count);
    }
  
    // Printing count of digits 0-9
    for (int i = 0; i <= 9; i++) 
    {
        cout << freq_count[i] <<  " ";
    }
}
// Driver code
int main()
{
    int x = 15, n = 3;
    countFrequency(x, n);
}
// This code is contributed by ihritik

Java

// Java implementation of above approach
import java.io.*;
import java.util.*;
public class GFG {
  
    // Function that traverses digits in a number and
    // modifies frequency count array
    static void countDigits(double val, long[] arr)
    {
        while ((long)val > 0) {
            long digit = (long)val % 10;
            arr[(int)digit]++;
            val = (long)val / 10;
        }
        return;
    }
  
    static void countFrequency(int x, int n)
    {
  
        // Array to keep count of digits
        long[] freq_count = new long[10];
  
        // Traversing through x^1 to x^n
        for (int i = 1; i <= n; i++) {
            // For power function, both its parameters are
            // to be in double
            double val = Math.pow((double)x, (double)i);
            // calling countDigits function on x^i
            countDigits(val, freq_count);
        }
  
        // Printing count of digits 0-9
        for (int i = 0; i <= 9; i++) {
            System.out.print(freq_count[i] + " ");
        }
    }
    // Driver code
    public static void main(String args[])
    {
        int x = 15, n = 3;
        countFrequency(x, n);
    }
}

Python 3

# Python 3 implementation 
# of above approach
import math
  
# Function that traverses digits 
# in a number and modifies 
# frequency count array
def countDigits(val, arr):
      
    while (val > 0) :
        digit = val % 10
        arr[int(digit)] += 1
        val = val // 10
          
    return;
  
def countFrequency(x, n):
      
    # Array to keep count of digits
    freq_count = [0] * 10
  
    # Traversing through x^1 to x^n
    for i in range(1, n + 1) :
          
        # For power function, 
        # both its parameters 
        # are to be in double
        val = math.pow(x, i)
          
        # calling countDigits 
        # function on x^i
        countDigits(val, freq_count)
          
    # Printing count of digits 0-9
    for i in range(10) :
        print(freq_count[i], end = " ");
  
# Driver code
if __name__ == "__main__":
      
    x = 15
    n = 3
    countFrequency(x, n)
  
# This code is contributed 
# by ChitraNayal

C#

// C# implementation of above approach
using System;
  
class GFG 
{
  
// Function that traverses digits 
// in a number and modifies 
// frequency count array
static void countDigits(double val, 
                        long[] arr)
{
    while ((long)val > 0)
    {
        long digit = (long)val % 10;
        arr[(int)digit]++;
        val = (long)val / 10;
    }
    return;
}
  
static void countFrequency(int x, int n)
{
  
    // Array to keep count of digits
    long[] freq_count = new long[10];
  
    // Traversing through x^1 to x^n
    for (int i = 1; i <= n; i++) 
    {
        // For power function, both its 
        // parameters are to be in double
        double val = Math.Pow((double)x,
                              (double)i);
                                
        // calling countDigits 
        // function on x^i
        countDigits(val, freq_count);
    }
  
    // Printing count of digits 0-9
    for (int i = 0; i <= 9; i++) 
    {
        Console.Write(freq_count[i] + " ");
    }
}
  
// Driver code
public static void Main()
{
    int x = 15, n = 3;
    countFrequency(x, n);
}
}
  
// This code is contributed 
// by Shashank

PHP

 0) 
    {
        $digit = $val % 10;
        $arr[(int)($digit)] += 1;
        $val = (int)($val / 10);
    }
    return;
}
  
function countFrequency($x, $n)
{
      
    // Array to keep count of digits
    $freq_count = array_fill(0, 10, 0);
  
    // Traversing through x^1 to x^n
    for ($i = 1; $i < $n + 1; $i++)
    {
          
        // For power function, 
        // both its parameters 
        // are to be in double
        $val = pow($x, $i);
          
        // calling countDigits 
        // function on x^i
        countDigits($val, $freq_count);
    } 
    // Printing count of digits 0-9
    for ($i = 0; $i < 10; $i++)
    {
        echo $freq_count[$i] . " ";
}
}
  
// Driver code
$x = 15;
$n = 3;
countFrequency($x, $n)
  
// This code is contributed by mits
?>

输出：

0 1 2 2 0 3 0 1 0 0